find the length width and area of an object where the perimeter is fixed at 20 meters
What kind of object?
a rectangle
P = 2L + 2W
30 = 2(5) + 2(5)
or
20 = 2(4) + 2(6)
or
20 = 2(3) + 2(7)
or
20 = 2(2) + 2(8)
or
20 = 2(1) + 2(9)
Correction:
the first combination should be:
20 = 2(5) + 2(5)
To find the length, width, and area of an object with a fixed perimeter, we need to follow these steps:
Step 1: Recall that the perimeter of a rectangle is the sum of the lengths of all its sides. In this case, since the perimeter is fixed at 20 meters, we can write the equation:
2(length + width) = 20
Step 2: Simplify the equation:
length + width = 10
Step 3: Now, we can express one variable in terms of the other. Let's solve for length in terms of width, by rearranging the equation:
length = 10 - width
Step 4: Since we know the formula for the area of a rectangle is length multiplied by width, we can substitute the expression for length from Step 3 into the area formula:
area = (10 - width) * width
Now, let's calculate the length, width, and area using this equation:
If we assume the width is 4 meters, we can plug in the value into the equation:
area = (10 - 4) * 4
area = 6 * 4
area = 24 square meters
Thus, if the width is 4 meters, the length would be 10 - 4 = 6 meters, and the area would be 24 square meters.
Similarly, you can repeat this process for different values of the width to find different combinations of length, width, and area with the fixed perimeter of 20 meters.